A well insulated 3-m X 4-m X 6-m room initially at 7 degrees C is heated by the radiator of a steam heating system. The radiator
has a volume of 15 L and is filled with superheated vapor at 200 kPa and 200 degrees C. At this moment both the inlet and the exit valves to the radiator are closed. A 120-W fan is used to distribute the air to the room. The pressure of the steam is observed to drop to 100 kPa after 45 min as a result of heat transfer to the room. Assuming constant specific heats for air at room temperature, determine the average termperature of air in 45 min. Assume the air pressure in the room remains constant at 100 kPa.
1 answer:
Answer:
The average temperature of air in 45 minutes is 10.74°C
Explanation:
Given
Time t = 45 minutes
Initial Air Temperature, T1 = 7°C
Fan work W = 120W
Radiator Volume = V1 = 15L
Initial Steam Pressure = P1 = 200kPa
Initial Steam Temperature = Ts1 = 200°C
Final Steam Pressure = P2 = 100kPa
The radiator is our system.
We take the system as a close system.
The energy balance is given as follows
-Qout = ∆U
-Qout = m(u2 - u1)
Calculating u2
From steam table (see attachment below)
At P1 and Ts1
P1 = 200kPa, Ts1 = 200°C
u1 = 2654.6KJ/Kg
v1 = 1.08049 m³/Kg
At P2 = 100kPa, v1 = v2
vf = 0.001043m³, vg = 1.6941m³/kg uf = 417.4kj/kg ufg = 2088.2kj/kg
x2 = (v2 - vf)/(vg - vf)
x2 = (1.08049 - 0.001043)/(1.6941 - 0.001043)
x2 = 1.079447/1.693057
x2 = 0.63757
u2 = uf + x2ufg
u2 = 417.4 + 0.63757 * 2088.2
u2 = 417.4 + 1331.374
u2 = 1748.774
Solving for m
m = V1/v1
m = 15L/1.08049
m = 0.015/1.08049
m = 0.01388Kg
So, -Qout = m(u2 - u1) becomes
0.01388(1748.774 - 2654.6)
= 0.01388 * -905.826
= -12.572
-Qout = -12.572
So, Qout = 12.572KJ
The dimensions of the room is 3 * 4 * 5.
The mass and volume of air in the room is given by
Volume = 3 * 4 * 6 = 72m³
Mass = P * V/RT
Where R = Gas Constant = 0.287
T = T1 = 7°C = 273+ 7 = 280K
P = P2 = 100kPa
Mass = (100 * 72)/(0.287 * 280)
Mass = 7200/80.36
Mass = 89.597Kg
The amount of work is defined by
Work = Fan work W * Time t
Work = 120 * 45 minutes
Work = 120 * 45 * 60seconds
Work = 324000J
Work (fan, in) = 324KJ
Taking the air in the room as the system
The energy balance is as follows
∆Energy = E(in) - E(out)
E(in) = Q(out) + W(fan,in)
E(out) = W(out)
∆Energy = ∆U
So, we have
∆U = Q(out) + W(fan,in) - W(out)
∆U + W(out) = Q(out) + W(fan,in)
∆U + W(out) = 12.572 + 324
∆U + W(out) = 336.572
∆U + W(out) = ∆H (Change in heat energy)
This is given by mc(T2 - T1)
Where m = 89.597kg
c = specific heat of air at constant pressure = 1.005
T1 = 7°C
Substituting in these values, we have
336.572 = 89.597 *1.005 (T2 - 7)
=> 336.572 = 90.044985(T2 - 7)
T2 - 7 = 3.74
T2 = 3.74 + 7
T2 = 10.74°C
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